Complex Analysis - Maths KU

Important paragraph. Reread it
several times.
☞
6.4 Zeros and Poles 337
If a function f has a removable singularity at the point z = z0, then we
can always supply an appropriate definition for the value of f(z0) so that
f becomes analytic at z = z0.For instance, since the right-hand side of
(3) is 1 when we set z = 0, it makes sense to define f(0) = 1.Hence the
function f(z) = (sin z)/z, as given in (3), is now defined and continuous at
every complex number z.Indeed, f is also analytic at z = 0 because it is
represented by the Taylor series 1 − z 2 /3! + z 4 /5! −··· centered at 0 (a
Maclaurin series).
EXAMPLE 2 Poles and Essential Singularity
(a) Dividing the terms of sin z = z − z3 z5
+
3! 5! −··· by z2 shows that
sin z
=
z2 principal
part
����
1
z
− z z3
+
3! 5! −···
for 0 < |z| < ∞.From this series we see that a−1 �= 0 and so z = 0 is a
simple pole of the function f(z) = (sin z)/z 2 .In like manner, we see that
z = 0 is a pole of order 3 of the function f(z) = (sin z)/z 4 considered in
Example 1 of Section 6.3.
(b) In Example 3 of Section 6.3 we showed that the Laurent expansion of
f(z) =1/(z − 1) 2 (z − 3) valid for 0 < |z − 1| < 2was
principal part
� �� �
1 1 1 z − 1
f(z) = − − − −
2(z − 1) 2 4(z − 1) 8 16 −···.
Since a−2 = − 1
2 �= 0, we conclude that z = 1 is a pole of order 2.
(c) In Example 6 of Section 6.3 we see from (19) that the principal part of
the Laurent expansion of the function f(z) =e 3/z valid for 0 < |z| < ∞
contains an infinite number of nonzero terms.This shows that z =0is
an essential singularity of f.
Zeros Recall, a number z0 is zero of a function f if f(z0) =0.Wesay
that an analytic function f has a zero of order n at z = z0 if
z0 is a zero of f and of its first n−1 derivatives
� �� �
f(z0) =0, f ′ (z0) =0, f ′′ (z0) =0, ... , f (n−1) (z0) =0, but f (n) (z0) �= 0. (4)